我有一个6D向量,我需要检查每个元素的邻域(每个方向2个元素)。当然,当我在矢量的边界上时,检查分段故障中的引线。我能做的就是切换大量的箱子。有没有更好的方法来解决这个问题?我也想过像try-catch这样的东西。
答案 0 :(得分:1)
仍然太笨重但是有效:
#include <iostream>
#include <array>
#include <vector>
typedef std::vector<int> Vector1D;
typedef std::vector<Vector1D> Vector2D;
typedef std::vector<Vector2D> Vector3D;
typedef std::vector<Vector3D> Vector4D;
typedef std::vector<Vector4D> Vector5D;
typedef std::vector<Vector5D> Vector6D;
typedef std::array<size_t, 6> Path;
bool GetVectorPathElement(Vector6D const &vector6D, Path const &path, int &val)
{
size_t i = 0, k = path[i];
if (vector6D.size() > k)
{
Vector5D const &vector5D = vector6D[k];
k = path[++i];
if (vector5D.size() > k)
{
Vector4D const &vector4D = vector5D[k];
k = path[++i];
if (vector4D.size() > k)
{
Vector3D const &vector3D = vector4D[k];
k = path[++i];
if (vector3D.size() > k)
{
Vector2D const &vector2D = vector3D[k];
k = path[++i];
if (vector2D.size() > k)
{
Vector1D const &vector1D = vector2D[k];
k = path[++i];
if (vector1D.size() > k)
{
val = vector1D[k];
return true;
}
}
}
}
}
}
std::cout << "Invalid path " << k << " at index " << i << std::endl;
return false;
}
int main()
{
Vector1D vector1D = { 1,2,3,4,5,6 };
Vector2D vector2D = { vector1D, vector1D, vector1D, vector1D, vector1D };
Vector3D vector3D = { vector2D, vector2D, vector2D, vector2D };
Vector4D vector4D = { vector3D, vector3D, vector3D };
Vector5D vector5D = { vector4D, vector4D };
Vector6D vector6D = { vector5D };
Path path = { 0,0,2,1,4,5 };
int element;
if (GetVectorPathElement(vector6D, path, element))
{
std::cout << "Path: ";
for (auto i : path)
std::cout << i << " ";
std::cout << "\nElement value at destination: " << element << std::endl;
}
return 0;
}